The Matrix Inverse

Solving an equation like $5x = 20$ is straightforward in basic algebra. To isolate the unknown variable $x$, you multiply both sides by the multiplicative inverse of 5, which is $1/5$ or $5^{-1}$. This gives you $(5^{-1}) \cdot 5x = (5^{-1}) \cdot 20$, which simplifies to $1x = 4$, or just $x = 4$.

We can apply a very similar logic to the matrix equation $Ax = b$. We want to isolate the vector $x$. However, you cannot "divide" by a matrix. Instead, we need to find the matrix equivalent of a multiplicative inverse: the matrix inverse.

The inverse of a square matrix $A$ is denoted as $A^{-1}$. When you multiply a matrix by its inverse, you get the identity matrix, $I$. The identity matrix, which we will look at more closely in the next section, is the matrix equivalent of the number 1. It is a square matrix with 1s on the main diagonal and 0s everywhere else.

The defining property of the inverse is:

$$A A^{-1} = A^{-1} A = I$$

Using the Inverse to Find the Solution

With the matrix inverse $A^{-1}$, we have a direct method for solving our system of equations. Let's see how it works by applying it to our original problem, $Ax = b$.

1. Start with the equation:

   $$Ax = b$$

2. Pre-multiply both sides by the inverse, $A^{-1}$: It's important to multiply from the left side on both sides of the equation, because matrix multiplication is not commutative (meaning $AB$ is not always equal to $BA$).

   $$A^{-1} (Ax) = A^{-1} b$$

3. Group the matrices: Since matrix multiplication is associative, we can regroup the terms on the left.

   $$(A^{-1} A) x = A^{-1} b$$

4. Simplify using the inverse property: We know that $A^{-1} A$ simplifies to the identity matrix $I$.

   $$I x = A^{-1} b$$

5. Isolate x: Multiplying any vector or matrix by the identity matrix $I$ leaves it unchanged (just like multiplying a number by 1). Therefore, $Ix$ is simply $x$.

   $$x = A^{-1} b$$

This final equation gives us a direct formula to find the unknown vector $x$. If we can find the inverse of matrix $A$, we can solve the system by simply multiplying $A^{-1}$ by the vector $b$.

The diagram shows how the inverse, $A^{-1}$, is used to transform the problem $Ax = b$ into a direct solution for $x$.

When Does an Inverse Exist?

Before we get too excited, there is a major condition: not all matrices have an inverse. For a matrix $A$ to have an inverse, it must meet two conditions:

1. The matrix must be square. An inverse is only defined for matrices with the same number of rows and columns (e.g., 2x2, 3x3). This aligns with the idea that for a unique solution to a system of linear equations, you typically need as many equations as you have unknown variables.
2. The matrix must be non-singular. This is a new term that you will explore in the next section. For now, think of a singular matrix as one that collapses data. For example, it might transform a 2D plane into a single line. If a matrix does this, information is lost, and there is no way to reverse the transformation. A non-singular, or invertible, matrix does not lose information.

Let's look at a simple example. Consider the matrix $A$:

$$A = \begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix}$$

The inverse of this matrix (which can be found using methods we'll see later) is:

$$A^{-1} = \begin{bmatrix} 1.5 & -0.5 \\ -2 & 1 \end{bmatrix}$$

To verify that this is indeed the inverse, let's multiply them together and see if we get the identity matrix:

$$A A^{-1} = \begin{bmatrix} 2 & 1 \\ 4 & 3 \end{bmatrix} \begin{bmatrix} 1.5 & -0.5 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} (2 \cdot 1.5 + 1 \cdot -2) & (2 \cdot -0.5 + 1 \cdot 1) \\ (4 \cdot 1.5 + 3 \cdot -2) & (4 \cdot -0.5 + 3 \cdot 1) \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$$

It works. If we had a system of equations where $A$ was our coefficient matrix, we could use this $A^{-1}$ to find the solution.

The matrix inverse is a foundational tool for solving linear systems. But this raises two important questions: How do we know if a matrix is invertible without just trying to find its inverse? And how do we calculate the inverse or solve the system efficiently? We'll begin to answer these questions in the next section on determinants.